Math Problem Statement

Let f(x) be a periodic function of x with period 3 and f(x) = |2x − 1| + x for

−1 < x ≤ 2. Sketch the graph of the curve y = f(x) in the interval [−3, 4]

Solution

To graph y=f(x)y = f(x), we will follow these steps:

Step 1: Understanding the function and periodicity

  • The function f(x)f(x) is periodic with a period of 3, which means that the graph of f(x)f(x) repeats every 3 units.
  • For the interval 1<x2-1 < x \leq 2, the function is defined as: f(x)=2x1+xf(x) = |2x - 1| + x

We will first sketch this function on 1<x2-1 < x \leq 2, and then use periodicity to repeat the pattern over the interval [3,4][-3, 4].

Step 2: Analyzing f(x)=2x1+xf(x) = |2x - 1| + x

  • 2x1|2x - 1| changes behavior at x=12x = \frac{1}{2}, since that’s where the absolute value function shifts from negative to positive.
  • So, we can break down f(x)f(x) into two pieces:
    • For x12x \leq \frac{1}{2}, 2x1=(2x1)=2x+1|2x - 1| = -(2x - 1) = -2x + 1. f(x)=(2x+1)+x=x+1f(x) = (-2x + 1) + x = -x + 1
    • For x>12x > \frac{1}{2}, 2x1=2x1|2x - 1| = 2x - 1. f(x)=(2x1)+x=3x1f(x) = (2x - 1) + x = 3x - 1

Thus, f(x)f(x) is defined piecewise as:

\begin{cases} -x + 1 & \text{for } -1 < x \leq \frac{1}{2}, \\ 3x - 1 & \text{for } \frac{1}{2} < x \leq 2. \end{cases}$$ ### Step 3: Plotting the function for $$-1 < x \leq 2$$ Now, let’s calculate some points for $$f(x)$$ in the interval $$-1 < x \leq 2$$: - For $$x = -1$$, $$f(-1) = (-(-1) + 1) = 2$$. - For $$x = 0$$, $$f(0) = (-0 + 1) = 1$$. - For $$x = \frac{1}{2}$$, $$f\left(\frac{1}{2}\right) = -\frac{1}{2} + 1 = \frac{1}{2}$$. - For $$x = 1$$, $$f(1) = (3(1) - 1) = 2$$. - For $$x = 2$$, $$f(2) = (3(2) - 1) = 5$$. We now have the following points: - $$(-1, 2), (0, 1), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 2), (2, 5)$$. We can plot these points and connect them using straight lines. ### Step 4: Extending the graph using periodicity Since $$f(x)$$ is periodic with a period of 3, we repeat the pattern from $$-1 < x \leq 2$$ over the entire interval $$[-3, 4]$$. - Repeat the pattern for the interval $$[-4, -1]$$ and $$[2, 5]$$, ensuring that the graph is continuous where the pieces meet. #### Final Graph: The graph will consist of piecewise linear segments based on the pattern derived above, repeating every 3 units. Would you like me to generate a sketch of this graph? ### Follow-Up Questions 1. What would happen if the period of $$f(x)$$ were changed to 2 instead of 3? 2. How would the graph change if we altered the function to $$f(x) = |3x - 2| + x$$? 3. Can you calculate the values of $$f(x)$$ at $$x = 3$$ and $$x = -2$$? 4. How would adding a constant $$c$$ to the function (i.e., $$f(x) + c$$) affect the graph? 5. Can you describe the effect of reflecting the graph of $$f(x)$$ over the x-axis? ### Tip: When dealing with periodic functions, it's helpful to fully analyze the behavior of the function over one period first, then replicate that pattern across the desired interval.

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Math Problem Analysis

Mathematical Concepts

Periodic Functions
Absolute Value
Piecewise Functions

Formulas

f(x) = |2x - 1| + x
f(x) = -x + 1 for -1 < x ≤ 1/2
f(x) = 3x - 1 for 1/2 < x ≤ 2

Theorems

Periodicity of Functions

Suitable Grade Level

Grades 10-12